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I'm a student in electrical engineering in Belgium and I use a multiple feedback bandpass filter in one of my projects.

bandpass filter

I tried to find out the transfer function of this filter and this is what I found:

$$ \frac{V_o}{V_i}=-\left( j\omega C+ \frac{2}{R_2}+\frac{R_1+R_3}{R_1R_2R_3}\cdot\frac{1}{j\omega C} \right) ^{-1} $$

With this expression I made a Bode diagram:

bode diagram

Is that normal that the frequencies other than 1kHz are still amplified (not as much but still) ? Is my transfer function right ?

Thanks !

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    \$\begingroup\$ It seems a normal transfer function for a bandpass filter...you can't expect a rectangle, and you have -20 dB for a factor 2 frequency multiplication-division...it's compatible with a 2nd order filter \$\endgroup\$ – clabacchio♦ Mar 27 '12 at 16:54
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    \$\begingroup\$ I don't work in audio, but isn't 80 dB (I assume you mean 10,000x voltage gain, since you don't have any load, and therefore you have no power gain) a lot of gain to be trying to get from a single op-amp stage? \$\endgroup\$ – The Photon Mar 28 '12 at 2:47
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I finally found the right transfer function (thanks to slateraptor at eevblog): $$\frac{V_o}{V_i}=-(\frac{R_3}{R_1+R_3})(\frac{j\omega R_2C}{-\omega^2(\frac{R_1R_3}{R_1+R_3})C^2R_2+j\omega(\frac{R_1R_3}{R_1+R_3})2C+1}) $$

I actually just forgot a \$\frac{1}{R_1}\$ factor in my first function.

This produces a much more logical bode diagram where only the 1kHz frequency is amplified: bode diagram

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  • \$\begingroup\$ Note that being an active filter, both transfer functions are plausible in an ideal world; you can offset your Bode diagram acting on the resistors \$\endgroup\$ – clabacchio♦ Mar 28 '12 at 12:13

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